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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>Consider a particular example:</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_3.html ./knowl/eq2_3.html ./knowl/eq2_3.html ./knowl/eq2_2.html">
\begin{equation}
\frac{\textrm{d} y}{\textrm{d} x}+\frac{1}{2} y=\frac{3}{2}.\tag{2.1.3}
\end{equation}
</div>
<p class="continuation">We want to find the solution to <a href="" class="xref" data-knowl="./knowl/eq2_3.html" title="Equation 2.1.3">(2.1.3)</a>. One method is like this:</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_3.html ./knowl/eq2_3.html ./knowl/eq2_3.html ./knowl/eq2_2.html">
\begin{equation}
\begin{aligned}
&amp;\frac{\textrm{d} y}{\textrm{d} x}=-\frac{y-3}{2} \rightarrow \frac{1}{y-3} \frac{\textrm{d} y}{\textrm{d} x}=-\frac{1}{2}\rightarrow  \frac{\textrm{d}}{\textrm{d} x} \ln |y-3|=-\frac{1}{2},\\
&amp;\rightarrow \ln |y-3|=-\frac{1}{2} x+C_1 \rightarrow |y-3|=e^{C_1} e^{-\frac{1}{2}x}\\
&amp;\rightarrow y-3=\uuline{\pm e^{C_1}}~ e^{-\frac{1}{2}x} \rightarrow y=3+C e^{-\frac{1}{2}x},
\end{aligned}\tag{2.1.4}
\end{equation}
</div>
<p class="continuation">where <span class="process-math">\(C_1\)</span> is arbitrary constant and <span class="process-math">\(C\)</span> is arbitrary constant not equal to zero. The other method to derive the solution to (<a href="" class="xref" data-knowl="./knowl/eq2_3.html" title="Equation 2.1.3">(2.1.3)</a>) is as follows. Multiplying <span class="process-math">\(e^{\frac{1}{2} x}\)</span> on both sides of (<a href="" class="xref" data-knowl="./knowl/eq2_3.html" title="Equation 2.1.3">(2.1.3)</a>), one has:</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_3.html ./knowl/eq2_3.html ./knowl/eq2_3.html ./knowl/eq2_2.html">
\begin{equation}
\begin{aligned}
&amp; \uuline{e^{\frac{1}{2} x} y^{\prime}+\frac{1}{2} e^{\frac{1}{2} x} y}=\frac{3}{2} e^{\frac{1}{2} x} \rightarrow \frac{\textrm{d}}{\textrm{d} x}\left(e^{\frac{1}{2} x} y \right)=\frac{3}{2} e^{\frac{1}{2} x}\\
&amp; \rightarrow e^{\frac{1}{2} x} y =3 e^{\frac{1}{2} x} +C \rightarrow y=3+Ce^{-\frac{1}{2}x},
\end{aligned}\tag{2.1.5}
\end{equation}
</div>
<p class="continuation">where <span class="process-math">\(C\)</span> is an arbitrary constant and <span class="process-math">\(e^{\frac{1}{2} x}\)</span> is called the <dfn class="terminology">integrating factor</dfn>. Through the integrating factor, the resulting equation is readily integrable. For a general linear first order ODE, it can not be solved by the direct method and one may consider using the integrating factor. The difficulty lies in <dfn class="terminology">how to find the integration factor for general case (<a href="" class="xref" data-knowl="./knowl/eq2_2.html" title="Equation 2.1.2">(2.1.2)</a>)?</dfn></p>
<span class="incontext"><a href="sec2_1.html#p-17" class="internal">in-context</a></span>
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